The circle

Let $M := \bS^1$, regarded as $\bS^1 \isom \bR/\bZ$; that is to say, we parametrize the circle by the half-open interval $[0,1)$ rather than $[0,2\pi)$, say. Then $\sA = \Coo(\bS^1)$ can be identified with periodic smooth functions on $\bR$ with period $1$:

$$\sA \isom \set{f \in \Coo(\bR) : f(t + 1) \equiv f(t)}.$$

Since $\Cl(\bR) = \bC 1 \oplus \bC e_1$ as a $\bZ_2$-graded algebra, we see that $\sB = \sA$ in this case; and since $n = 1$, $m = 0$ and $2^m = 1$, there is a ``trivial'' spin structure given by $\sS := \sA$ itself. The charge conjugation is just $C = K$, where $K$ means complex conjugation of functions. With the flat metric on the circle, the Dirac operator is just

$$\Dslash := -i \frac{d}{dt}.$$

The point is that the closed span of these eigenvectors is all of $\sH$, so that $\spec(\Dslash)$ contains no more than the corresponding eigenvalues.

Next, consider

$$\sS' := \set{\phi \in \Coo(\bR) : \phi(t + 1) \equiv - \phi(t)},$$

which can be thought of as the space of smooth functions on the interval $[0,1]$ ``with antiperiodic boundary conditions''.

The circle $\bS^1$ thus carries two inequivalent spin structures: their inequivalence is most clearly manifest in the different spectra of the Dirac operators. Notice that $0 \in \spec(\Dslash)$ for the ``untwisted'' spin structure where $\sS = \sA$, while $0 \notin \spec(\Dslash)$ for the ``twisted'' spin structure whose spinor module is $\sS'$. There are no more spin structures to be found, since $H^1(\bS^1,\bZ_2) = \bZ_2$.